Introduction

(a) Show that $\varphi \left(\mathbf{x}\right)\mathbf{=}{\mathbf{x}}^{\mathbf{2}}$ is an explicit solution to $\mathbf{x}\frac{\mathbf{dy}}{\mathbf{dx}}\mathbf{=}\mathbf{2}\mathbf{y}$ on the interval $\left(-\infty ,\infty \right)$.

(b) Show that $\varphi \mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}{\mathbf{e}}^{\mathbf{x}}\mathbf{-}\mathbf{x}$, is an explicit solution to $\frac{\mathbf{dy}}{\mathbf{dx}}\mathbf{+}{\mathbf{y}}^{\mathbf{2}}\mathbf{=}{\mathbf{e}}^{\mathbf{x}}\mathbf{+}\left(\mathbf{1}\mathbf{-}\mathbf{2}\mathbf{x}\right){\mathbf{e}}^{\mathbf{x}}\mathbf{+}{\mathbf{x}}^{\mathbf{2}}\mathbf{-}\mathbf{1}$ on the interval $\left(-\infty ,\infty \right)$.

(c) Show that $\varphi \left(\mathbf{x}\right)\mathbf{=}{\mathbf{x}}^{\mathbf{2}}\mathbf{-}{\mathbf{x}}^{\mathbf{-}\mathbf{1}}$ is an explicit solution to ${\mathbf{x}}^{\mathbf{2}}\frac{{\mathbf{d}}^{\mathbf{2}}\mathbf{y}}{{\mathbf{dx}}^{\mathbf{2}}}\mathbf{=}\mathbf{2}\mathbf{y}$ on the interval $\left(0,\infty \right)$.

In problems 1-4 Use Euler’s method to approximate the solution to the given initial value problem at the points $\mathbf{x}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{1}\mathbf{,}\mathbf{0}\mathbf{.}\mathbf{2}\mathbf{,}\mathbf{0}\mathbf{.}\mathbf{3}\mathbf{,}\mathbf{0}\mathbf{.}\mathbf{4}$ , and $\mathbf{0}\mathbf{.}\mathbf{5}$ , using steps of size $\mathbf{0}\mathbf{.}\mathbf{1}\left(\mathbf{h}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{1}\right)$ .

$\frac{\mathbf{dy}}{\mathbf{dx}}\mathbf{=}\frac{\mathbf{-}\mathbf{x}}{\mathbf{y}}$ , $\mathbf{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{4}$

The directional field for $\frac{\mathbf{dy}}{\mathbf{dx}}\mathbf{=}\frac{\mathbf{4}\mathbf{x}}{\mathbf{y}}$ in shown in figure 1.12. 

(a) Verify that the straight lines $\mathbf{y}\mathbf{=}\mathbf{\pm } \mathbf{2}\mathbf{x}$ are solution curves, provided $\mathbf{x}\mathbf{\ne }\mathbf{0}$.

(b) Sketch the solution curve with initial condition y (0) = 2.

(c) Sketch the solution curve with initial condition y(2) = 1.

(d) What can you say about the behaviour of the above solution as $\mathbf{x}\mathbf{\to }\mathbf{+}\mathbf{\infty }$? How about $\mathbf{x}\mathbf{\to }\mathbf{-}\mathbf{\infty }$?

In problems 1-6, identify the independent variable, dependent variable, and determine whether the equation is linear or nonlinear.

$\mathbf{5}\frac{\mathbf{dx}}{\mathbf{dt}}\mathbf{+}\mathbf{5}{\mathbf{x}}^{\mathbf{2}}\mathbf{+}\mathbf{3}\mathbf{=}\mathbf{0}$

In Problems , identify the equation as separable, linear, exact, or having an integrating factor that is a function of either x alone or y alone.

$\left(\mathbf{2}\mathbf{x}\mathbf{+}{\mathbf{yx}}^{\mathbf{-}\mathbf{1}}\right)\mathbf{dx}\mathbf{+}\left(\mathbf{xy}\mathbf{-}\mathbf{1}\right)\mathbf{dy}\mathbf{=}\mathbf{0}$

In problems  $\mathbf{1}\mathbf{-}\mathbf{4}$ Use Euler’s method to approximate the solution to the given initial value problem at the points x = 0.1, 0.2, 0.3, 0.4, and 0.5, using steps of size 0.1 (h = 0.1).

$\frac{\mathbf{dy}}{\mathbf{dx}}\mathbf{=}\mathbf{y}\mathbf{(}\mathbf{2}\mathbf{-}\mathbf{y}\mathbf{)}$, $\mathbf{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{3}$

The direction field for $\frac{\mathbf{dy}}{\mathbf{dx}}\mathbf{=}\mathbf{2}\mathbf{x}\mathbf{+}\mathbf{y}$ as shown in figure 1.13.

1. Sketch the solution curve that passes through (0, -2). From this sketch, write the equation for the solution.

b. Sketch the solution curve that passes through (-1, 3).

c. What can you say about the solution in part (b) as $\mathbf{x}\mathbf{\to }\mathbf{+}\mathbf{\infty }$? How about $\mathbf{x}\mathbf{\to }\mathbf{-}\mathbf{\infty }$?

In problems 1-6, identify the independent variable, dependent variable, and determine whether the equation is linear or nonlinear.

$\mathbf{3}\mathit{r}\mathbf{-}\mathit{c}\mathit{o}\mathit{s}\mathit{\theta }\frac{\mathbf{d}\mathbf{r}}{\mathbf{d}\mathbf{\theta }}\mathbf{=}\mathit{s}\mathit{i}\mathit{n}\mathit{\theta }$

In problems  $\mathbf{1}\mathbf{-}\mathbf{4}$ Use Euler’s method to approximate the solution to the given initial value problem at the points x = 0.1, 0.2, 0.3, 0.4, and 0.5, using steps of size 0.1 (h = 0.1).

$\frac{\mathbf{dy}}{\mathbf{dx}}\mathbf{=}\mathbf{x}\mathbf{+}\mathbf{y}\mathbf{,}\mathbf{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{1}$

A model for the velocity v at time t of a certain object falling under the influence of gravity in a viscous medium is given by the equation $\frac{\mathbf{dv}}{\mathbf{dt}}\mathbf{=}\mathbf{1}\mathbf{-}\frac{\mathbf{v}}{\mathbf{8}}$.  From the direction field shown in Figure 1.14, sketch the solutions with the initial conditions v(0) = 5, 8, and 15. Why is the value v = 8 called the “terminal velocity”?

                                                           Figure 1.14

In problems 1-6, identify the independent variable, dependent variable, and determine whether the equation is linear or nonlinear.

${\mathbf{y}}^{\mathbf{3}}\frac{{\mathbf{d}}^{\mathbf{2}}\mathbf{x}}{{\mathbf{dy}}^{\mathbf{2}}}\mathbf{+}\mathbf{3}\mathbf{x}\mathbf{-}\frac{\mathbf{8}}{\mathbf{y}\mathbf{-}\mathbf{1}}\mathbf{=}\mathbf{0}$

Use the improved Euler’s method subroutine with step size h = 0.1 to approximate the solution to $\mathbf{y}\mathbf{\text{'}}\mathbf{=}\mathbf{4}\mathbf{cos}\mathbf{(}\mathbf{x}\mathbf{+}\mathbf{y}\mathbf{)}\mathbf{,}\mathbf{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}$, at the points $\mathbf{x}\mathbf{=}\mathbf{0}\mathbf{,} \mathbf{0}\mathbf{.}\mathbf{1}\mathbf{,} \mathbf{0}\mathbf{.}\mathbf{2}\mathbf{,}...\mathbf{,} \mathbf{1}\mathbf{.}\mathbf{0}$. Use your answers to make a rough sketch of the solution on [0,1].

In problems 1-4 Use Euler’s method to approximate the solution to the given initial value problem at the points $\mathbf{x}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{1}\mathbf{,}\mathbf{0}\mathbf{.}\mathbf{2}\mathbf{,}\mathbf{0}\mathbf{.}\mathbf{3}\mathbf{,}\mathbf{0}\mathbf{.}\mathbf{4}$ , and $\mathbf{0}\mathbf{.}\mathbf{5}$ , using steps of size $\mathbf{0}\mathbf{.}\mathbf{1}\left(\mathbf{h}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{1}\right)$ .

$\frac{\mathbf{d}\mathbf{y}}{\mathbf{d}\mathbf{x}}\mathbf{=}\frac{\mathbf{x}}{\mathbf{y}}\mathbf{ }\mathbf{ }\mathbf{,}\mathbf{ }\mathbf{ }\mathit{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{-}\mathbf{1}$

In problems 1-6, identify the independent variable, dependent variable, and determine whether the equation is linear or nonlinear.

${\mathbf{t}}^{\mathbf{3}}{\left(\frac{{\mathbf{d}}^{\mathbf{3}}\mathbf{x}}{{\mathbf{dt}}^{\mathbf{3}}}\right)}^{\mathbf{2}}\mathbf{+}\mathbf{3}\mathbf{t}\mathbf{-}\mathbf{x}\mathbf{=}\mathbf{0}$

Find a general solution for the given differential equation.

(a) ${\mathit{y}}^{\mathbf{\left(}\mathbf{4}\mathbf{\right)}}\mathbf{+}\mathbf{2}{\mathit{y}}^{\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\text{'}}}\mathbf{-}\mathbf{4}{\mathit{y}}^{\mathbf{\text{'}}\mathbf{\text{'}}}\mathbf{-}\mathbf{2}{\mathit{y}}^{\mathbf{\text{'}}}\mathbf{+}\mathbf{3}\mathit{y}\mathbf{=}\mathbf{0}$

(b) ${\mathit{y}}^{\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\text{'}}}\mathbf{+}\mathbf{3}{\mathit{y}}^{\mathbf{\text{'}}\mathbf{\text{'}}}\mathbf{-}\mathbf{5}{\mathit{y}}^{\mathbf{\text{'}}}\mathbf{+}\mathit{y}\mathbf{=}\mathbf{0}$

(c) ${\mathit{y}}^{\mathbf{\left(}\mathbf{5}\mathbf{\right)}}\mathbf{-}{\mathit{y}}^{\mathbf{\left(}\mathbf{4}\mathbf{\right)}}\mathbf{+}\mathbf{2}{\mathit{y}}^{\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\text{'}}}\mathbf{-}\mathbf{2}{\mathit{y}}^{\mathbf{\text{'}}\mathbf{\text{'}}}\mathbf{+}{\mathit{y}}^{\mathbf{\text{'}}}\mathbf{-}\mathit{y}\mathbf{=}\mathbf{0}$

(d) ${\mathit{y}}^{\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\text{'}}}\mathbf{-}\mathbf{2}{\mathit{y}}^{\mathbf{\text{'}}\mathbf{\text{'}}}\mathbf{-}{\mathit{y}}^{\mathbf{\text{'}}}\mathbf{+}\mathbf{2}\mathit{y}\mathbf{=}{\mathit{e}}^{\mathbf{x}}\mathbf{+}\mathit{x}$

The logistic equation for the population (in thousands) of a certain species is given by $\frac{\mathbf{dp}}{\mathbf{dt}}\mathbf{=}\mathbf{3}\mathbf{p}\mathbf{-}\mathbf{2}{\mathbf{p}}^{\mathbf{2}}$ .

⦁    Sketch the direction field by using either a computer software package or the method of isoclines.

⦁    If the initial population is 3000 [that is, p(0) = 3], what can you say about the limiting population?

⦁    If $\mathbf{p}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{8}$ , what is ${\mathbf{lim}}_{\mathbf{t}\to \mathbf{+}\infty }\mathbf{p}\mathbf{\left(}\mathbf{t}\mathbf{\right)}$ ?

⦁    Can a population of 2000 ever decline to 800?

In Problems 1 and 2, verify that the pair x(t), and y(t) is a solution to the given system. Sketch the trajectory of the given solution in the phase plane.

$$\begin{gathered} \frac{\mathbf{dx}}{\mathbf{dt}}\mathbf{=}\mathbf{1}\mathbf{,}\frac{\mathbf{dy}}{\mathbf{dt}}\mathbf{=}\mathbf{3}{\mathbf{x}}^{\mathbf{2}}\mathbf{;} \\ \mathbf{x}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}\mathbf{t}\mathbf{+}\mathbf{1}\mathbf{,}\mathbf{y}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}{\mathbf{t}}^{\mathbf{3}}\mathbf{+}\mathbf{3}{\mathbf{t}}^{\mathbf{2}}\mathbf{+}\mathbf{3}\mathbf{t} \end{gathered}$$

Use Euler’s method with step size h = 0.1 to approximate the solution to the initial value problem 

 $\mathbf{y}\mathbf{\text{'}}\mathbf{=}\mathbf{x}\mathbf{-}{\mathbf{y}}^{\mathbf{2}}$, y (1) = 0 at the points $\mathbf{x}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{1}\mathbf{,} \mathbf{1}\mathbf{.}\mathbf{2}\mathbf{,} \mathbf{1}\mathbf{.}\mathbf{3}\mathbf{,} \mathbf{1}\mathbf{.}\mathbf{4} \text{and} \mathbf{1}\mathbf{.}\mathbf{5}$ .

Find a general solution for the differential equation with x as the independent variable:

$y^m+3y^n+28y\text{'}+26y=0$

In Problems 10–13, use the vectorized Euler method with h = 0.25 to find an approximation for the solution to the given initial value problem on the specified interval.

$\mathbf{(}\mathbf{1}\mathbf{+}{\mathbf{t}}^{\mathbf{2}}\mathbf{)}\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{y}\mathbf{\text{'}}\mathbf{-}\mathbf{y}\mathbf{=}\mathbf{0}\mathbf{;}\mathbf{y}\mathbf{(}\mathbf{0}\mathbf{)}\mathbf{=}\mathbf{1}\mathbf{,}\mathbf{y}\mathbf{\text{'}}\mathbf{(}\mathbf{0}\mathbf{)}\mathbf{=}\mathbf{-}\mathbf{1}\mathbf{on}\mathbf{[}\mathbf{0}\mathbf{,}\mathbf{1}\mathbf{]}$

In Problems 10–13, use the vectorized Euler method with h = 0.25 to find an approximation for the solution to the given initial value problem on the specified interval.

${\mathbf{t}}^{\mathbf{2}}\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{y}\mathbf{=}\mathbf{t}\mathbf{+}\mathbf{2}\mathbf{;}\mathbf{y}\mathbf{\left(}\mathbf{1}\mathbf{\right)}\mathbf{=}\mathbf{1}\mathbf{,}\mathbf{y}\mathbf{\text{'}}\mathbf{\left(}\mathbf{1}\mathbf{\right)}\mathbf{=}\mathbf{-}\mathbf{1}\mathbf{on}\mathbf{[}\mathbf{1}\mathbf{,}\mathbf{2}\mathbf{]}$

In Problems 10–13, use the vectorized Euler method with h = 0.25 to find an approximation for the solution to the given initial value problem on the specified interval.

$\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{=}{\mathbf{t}}^{\mathbf{2}}\mathbf{-}{\mathbf{y}}^{\mathbf{2}}\mathbf{;}\mathbf{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}\mathbf{,}\mathbf{y}\mathbf{\text{'}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{1}\mathbf{on}\mathbf{[}\mathbf{0}\mathbf{,}\mathbf{1}\mathbf{]}$

In Problems 14–24, you will need a computer and a programmed version of the vectorized classical fourth-order Runge–Kutta algorithm. (At the instructor’s discretion, other algorithms may be used.)†

Using the vectorized Runge–Kutta algorithm with h = 0.5, approximate the solution to the initial value $\mathbf{3}{\mathbf{t}}^{\mathbf{2}}\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}\mathbf{5}\mathbf{ty}\mathbf{\text{'}}\mathbf{+}\mathbf{5}\mathbf{y}\mathbf{=}\mathbf{0}\mathbf{;}\mathbf{y}\mathbf{\left(}\mathbf{1}\mathbf{\right)}\mathbf{=}\mathbf{0}\mathbf{,}\mathbf{y}\mathbf{\text{'}}\mathbf{\left(}\mathbf{1}\mathbf{\right)}\mathbf{=}\frac{\mathbf{2}}{\mathbf{3}}$ problemat t = 8.

Compare this approximation to the actual solution $\mathbf{y}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}{\mathbf{t}}^{\frac{\mathbf{5}}{\mathbf{3}}}\mathbf{-}\mathbf{t}$.

In Problems 14–24, you will need a computer and a programmed version of the vectorized classical fourth-order Runge–Kutta algorithm. (At the instructor’s discretion, other algorithms may be used.)†

Using the vectorized Runge–Kutta algorithm, approximate the solution to the initial value problem $\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{=}{\mathbf{t}}^{\mathbf{2}}\mathbf{+}{\mathbf{y}}^{\mathbf{2}}\mathbf{;}\mathbf{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{1}\mathbf{,}\mathbf{y}\mathbf{\text{'}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}$ at  $\mathbf{t}\mathbf{=}\mathbf{1}$. Starting with $\mathbf{h}\mathbf{=}\mathbf{1}$, continue halving the step size until two successive approximations of both $\mathbf{y}\left(\mathbf{1}\right)$ and $\mathbf{y}\mathbf{\text{'}}\left(\mathbf{1}\right)$ differ by at most 0.1.

In Problems 14–24, you will need a computer and a programmed version of the vectorized classical fourth-order Runge–Kutta algorithm. (At the instructor’s discretion, other algorithms may be used.)†

Using the vectorized Runge–Kutta algorithm for systems with $\mathit{h}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{175}$, approximate the solution to the initial value problem $\mathbf{x}\mathbf{\text{'}}\mathbf{=}\mathbf{2}\mathbf{x}\mathbf{-}\mathbf{y}\mathbf{;}\mathbf{x}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}\mathbf{,}\mathbf{y}\mathbf{\text{'}}\mathbf{=}\mathbf{3}\mathbf{x}\mathbf{+}\mathbf{6}\mathbf{y}\mathbf{;}\mathbf{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{-}\mathbf{2}$ at $\mathbf{t}\mathbf{=}\mathbf{1}$. 

Compare this approximation to the actual solution.

In Problems 14–24, you will need a computer and a programmed version of the vectorized classical fourth-order Runge–Kutta algorithm. (At the instructor’s discretion, other algorithms may be used.)†

Using the vectorized Runge–Kutta algorithm, approximate the solution to the initial value problem

$$\begin{gathered} \frac{\mathbf{du}}{\mathbf{dx}}\mathbf{=}\mathbf{3}\mathbf{u}\mathbf{-}\mathbf{4}\mathbf{v}\mathbf{;}\mathbf{u}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{1}\mathbf{\text{'}} \\ \frac{\mathbf{dv}}{\mathbf{dx}}\mathbf{=}\mathbf{2}\mathbf{u}\mathbf{-}\mathbf{3}\mathbf{v}\mathbf{;}\mathbf{v}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{1} \end{gathered}$$

at x = 1. Starting with h=1, continue halving the step size until two successive approximations of u(1)and v(1) differ by at most 0.001.

Combat Model. A simplified mathematical model for conventional versus guerrilla combat is given by the system $\mathbf{x}{\mathbf{\text{'}}}_{\mathbf{1}}\mathbf{=}\mathbf{-}\mathbf{(}\mathbf{0}\mathbf{.}\mathbf{1}\mathbf{)}{\mathbf{x}}_{\mathbf{1}}{\mathbf{x}}_{\mathbf{2}}\mathbf{;} {\mathbf{x}}_{\mathbf{1}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{10}\mathbf{;} \mathbf{x}{\mathbf{\text{'}}}_{\mathbf{2}}\mathbf{=}\mathbf{-}{\mathbf{x}}_{\mathbf{1}}\mathbf{;}{\mathbf{x}}_{\mathbf{2}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{15}$ where ${\mathbf{x}}_{\mathbf{1}}$ and ${\mathbf{x}}_{\mathbf{2}}$ are the strengths of guerrilla and conventional troops, respectively, and 0.1 and 1 are the combat effectiveness coefficients. Who will win the conflict: the conventional troops or the guerrillas? [Hint: Use the vectorized Runge–Kutta algorithm for systems with h=0.1 to approximate the solutions.]

In Project C of Chapter 4, it was shown that the simple pendulum equation $\mathbf{\theta }\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{+}\mathbf{sin\theta }\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}\mathbf{0}$ has periodic solutions when the initial displacement and velocity show that the period of the solution may depend on the initial conditions by using the vectorized Runge–Kutta algorithm with h = 0.02 to approximate the solutions to the simple pendulum problem on

[0, 4] for the initial conditions:

$$\begin{gathered} \mathbf{(}\mathbf{a}\mathbf{)}\mathbf{ }\mathbf{ }\mathbf{\theta }\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{1}\mathbf{,}\mathbf{\theta }\mathbf{\text{'}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0} \\ \mathbf{(}\mathbf{b}\mathbf{)}\mathbf{ }\mathbf{ }\mathbf{\theta }\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{5}\mathbf{,}\mathbf{\theta }\mathbf{\text{'}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0} \\ \mathbf{(}\mathbf{c}\mathbf{)}\mathbf{ }\mathbf{ }\mathbf{\theta }\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{0}\mathbf{,}\mathbf{\theta }\mathbf{\text{'}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0} \end{gathered}$$

[Hint: Approximate the length of time it takes to reach].

Oscillations and Nonlinear Equations. For the initial value problem $\mathbf{x}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{(}\mathbf{0}\mathbf{.}\mathbf{1}\mathbf{\left)}\mathbf{\right(}\mathbf{1}\mathbf{-}{\mathbf{x}}^{\mathbf{2}}\mathbf{)}\mathbf{x}\mathbf{\text{'}}\mathbf{+}\mathbf{x}\mathbf{=}\mathbf{0}\mathbf{;}\mathbf{x}\mathbf{(}\mathbf{0}\mathbf{)}\mathbf{=}{\mathbf{x}}_{\mathbf{o}}\mathbf{,}\mathbf{x}\mathbf{\text{'}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}$ using the vectorized Runge–Kutta algorithm with h = 0.02 to illustrate that as t increases from 0 to 20, the solution x exhibits damped oscillations when ${\mathbf{x}}_{\mathbf{o}}\mathbf{=}\mathbf{1}$, whereas exhibits expanding oscillations when ${\mathbf{x}}_{\mathbf{o}}\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{1}\mathbf{,}$.

Nonlinear Spring. The Duffing equation $\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{y}\mathbf{+}{\mathbf{ry}}^{\mathbf{3}}\mathbf{=}\mathbf{0}$ where r is a constant is a model for the vibrations of amass attached to a nonlinear spring. For this model, does the period of vibration vary as the parameter r is varied?

Does the period vary as the initial conditions are varied? [Hint: Use the vectorized Runge–Kutta algorithm with h = 0.1 to approximate the solutions for r = 1 and 2,

with initial conditions $\mathbf{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{a}\mathbf{,}\mathbf{y}\mathbf{\text{'}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}$ for a = 1, 2, and 3.]

Pendulum with Varying Length. A pendulum is formed by a mass m attached to the end of a wire that is attached to the ceiling. Assume that the length l(t)of the wire varies with time in some predetermined fashion. If

U(t) is the angle in radians between the pendulum and the vertical, then the motion of the pendulum is governed for small angles by the initial value problem ${\mathbf{l}}^{\mathbf{2}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{\theta }\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{+}\mathbf{2}\mathbf{l}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{l}\mathbf{\text{'}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{\theta }\mathbf{\text{'}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{+}\mathbf{gl}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{sin}\mathbf{\left(}\mathbf{\theta }\mathbf{\right(}\mathbf{t}\mathbf{\left)}\mathbf{\right)}\mathbf{=}\mathbf{0}\mathbf{;}\mathbf{\theta }\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}{\mathbf{\theta }}_{\mathbf{o}}\mathbf{,}\mathbf{\theta }\mathbf{\text{'}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}{\mathbf{\theta }}_{\mathbf{1}}$ where g is the acceleration due to gravity. Assume that $\mathbf{l}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}{\mathbf{l}}_{\mathbf{o}}\mathbf{+}{\mathbf{l}}_{\mathbf{1}}\mathbf{cos}\mathbf{(}\mathbf{\omega t}\mathbf{-}\varphi \mathbf{)}$ where ${\mathbf{l}}_1$ is much smaller than ${\mathbf{l}}_{\mathbf{o}}$. (This might be a model for a person on a swing, where the pumping action changes the distance from the center of mass of the swing to the point where the swing is attached.) To simplify the computations, take g = 1. Using the Runge– Kutta algorithm with h = 0.1, study the motion of the pendulum when ${\mathbf{\theta }}_{\mathbf{o}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{05}\mathbf{,}{\mathbf{\theta }}_{\mathbf{1}}\mathbf{=}\mathbf{0}\mathbf{,}{\mathbf{l}}_{\mathbf{o}}\mathbf{=}\mathbf{1}\mathbf{,}{\mathbf{l}}_{\mathbf{1}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{1}\mathbf{,}\mathbf{\omega }\mathbf{=}\mathbf{1}\mathbf{,}\mathit{\varphi }\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{02}$. In particular, does the pendulum ever attain an angle greater in absolute value than the initial angle ${\mathbf{\theta }}_{\mathbf{o}}$?

Using the Runge–Kutta algorithm for systems with h = 0.05, approximate the solution to the initial value problem $\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}{\mathbf{y}}^{\mathbf{2}}\mathbf{=}\mathbf{t}\mathbf{;}\mathbf{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{1}\mathbf{,}\mathbf{y}\mathbf{\text{'}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}\mathbf{,}\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{1}$ at t=1.

Lunar Orbit. The motion of a moon moving in a planar orbit about a planet is governed by the equations $\frac{{\mathbf{d}}^{\mathbf{2}}\mathbf{x}}{{\mathbf{dt}}^{\mathbf{2}}}\mathbf{=}\mathbf{-}\mathbf{G}\frac{\mathbf{mx}}{{\mathbf{r}}^{\mathbf{3}}}\mathbf{,}\frac{{\mathbf{d}}^{\mathbf{2}}\mathbf{y}}{{\mathbf{dt}}^{\mathbf{2}}}\mathbf{=}\mathbf{-}\mathbf{G}\frac{\mathbf{my}}{{\mathbf{r}}^{\mathbf{3}}}$ where $\mathbf{r}\mathbf{=}\mathbf{(}{\mathbf{x}}^{\mathbf{2}}\mathbf{+}{\mathbf{y}}^{\mathbf{2}}{\mathbf{)}}^{\frac{\mathbf{1}}{\mathbf{2}}}$, G is the gravitational constant, and m is the mass of the planet. Assume Gm = 1. When  $\mathbf{x}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{1}\mathbf{,}\mathbf{x}\mathbf{\text{'}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}\mathbf{,}\mathbf{y}\mathbf{\text{'}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{1}$ the motion is a circular orbit of radius 1 and period $\mathbf{2}\mathbf{\pi }$.

(a) The setting ${\mathbf{x}}_{\mathbf{1}}\mathbf{=}\mathbf{x}\mathbf{,}{\mathbf{x}}_{\mathbf{2}}\mathbf{=}\mathbf{x}\mathbf{\text{'}}\mathbf{,}{\mathbf{x}}_{\mathbf{3}}\mathbf{=}\mathbf{y}\mathbf{,}{\mathbf{x}}_{\mathbf{4}}\mathbf{=}\mathbf{y}\mathbf{\text{'}}$ expresses the governing equations as a first-order system in normal form.

(b) Using $\mathbf{h}\mathbf{=}\frac{\mathbf{2}\mathbf{\pi }}{\mathbf{100}}\mathbf{\approx }\mathbf{0}\mathbf{.}\mathbf{0628318}$, compute one orbit of this moon (i.e., do N = 100 steps.). Do your approximations agree with the fact that the orbit is a circle of radius 1?

Competing Species. Let pi(t) denote, respectively, the populations of three competing species ${\mathbf{S}}_{\mathbf{i}}\mathbf{,}\mathbf{i}\mathbf{=}\mathbf{1}\mathbf{,}\mathbf{2}\mathbf{,}\mathbf{3}\mathbf{.}$Suppose these species have the same growth rates, and the maximum population that the habitat can support is the same for each species. (We assume it to be one unit.) Also, suppose the competitive advantage that ${\mathbf{S}}_{\mathbf{1}}$ has over ${\mathbf{S}}_{\mathbf{2}}$ is the same as that of ${\mathbf{S}}_{\mathbf{2}}$ over ${\mathbf{S}}_{\mathbf{3}}$ and over. This situation is modeled by the system

$$\begin{gathered} \mathbf{p}{\mathbf{\text{'}}}_{\mathbf{1}}\mathbf{=}{\mathbf{p}}_{\mathbf{1}}\mathbf{(}\mathbf{1}\mathbf{-}{\mathbf{p}}_{\mathbf{1}}\mathbf{-}{\mathbf{ap}}_{\mathbf{2}}\mathbf{-}{\mathbf{bp}}_{\mathbf{3}}\mathbf{)} \\ \mathbf{p}{\mathbf{\text{'}}}_{\mathbf{2}}\mathbf{=}{\mathbf{p}}_{\mathbf{2}}\mathbf{(}\mathbf{1}\mathbf{-}\mathbf{b}{\mathbf{p}}_{\mathbf{1}}\mathbf{-}{\mathbf{p}}_{\mathbf{2}}\mathbf{-}{\mathbf{ap}}_{\mathbf{3}}\mathbf{)} \\ \mathbf{p}{\mathbf{\text{'}}}_{\mathbf{3}}\mathbf{=}{\mathbf{p}}_{\mathbf{3}}\mathbf{(}\mathbf{1}\mathbf{-}\mathbf{a}{\mathbf{p}}_{\mathbf{1}}\mathbf{-}{\mathbf{bp}}_{\mathbf{2}}\mathbf{-}{\mathbf{p}}_{\mathbf{3}}\mathbf{)} \end{gathered}$$

where a and b are positive constants. To demonstrate the population dynamics of this system when a = b = 0.5, use the Runge–Kutta algorithm for systems with h = 0.1 to approximate the populations over the time interval [0, 10] under each of the following initial conditions:

$$\begin{gathered} \mathbf{(}\mathbf{a}\mathbf{)}\mathbf{ }{\mathbf{p}}_{\mathbf{1}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{0}\mathbf{,}{\mathbf{p}}_{\mathbf{2}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{1}\mathbf{,}{\mathbf{p}}_{\mathbf{3}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{1} \\ \mathbf{(}\mathbf{b}\mathbf{)}\mathbf{ }{\mathbf{p}}_{\mathbf{1}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{1}\mathbf{,}{\mathbf{p}}_{\mathbf{2}}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{0}\mathbf{,}{\mathbf{p}}_{\mathbf{3}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{1} \\ \mathbf{(}\mathbf{c}\mathbf{)}\mathbf{ }{\mathbf{p}}_{\mathbf{1}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{1}\mathbf{,}{\mathbf{p}}_{\mathbf{2}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{1}\mathbf{,}{\mathbf{p}}_{\mathbf{3}}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{0} \end{gathered}$$

Spring Pendulum. Let a mass be attached to one end of a spring with spring constant k and the other end attached to the ceiling. Let ${\mathbf{l}}_{\mathbf{o}}$ be the natural length of the spring, and let l(t) be its length at time t. If $\mathbf{\theta }\mathbf{\left(}\mathbf{t}\mathbf{\right)}$ is the angle between the pendulum and the vertical, then the motion of the spring pendulum is governed by the system 

$$\begin{gathered} \mathbf{l}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{-}\mathbf{l}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{\theta }\mathbf{\text{'}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{-}\mathbf{gcos\theta }\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{+}\frac{\mathbf{k}}{\mathbf{m}}\mathbf{(}\mathbf{l}\mathbf{-}{\mathbf{l}}_{\mathbf{o}}\mathbf{)}\mathbf{=}\mathbf{0} \\ {\mathbf{l}}^{\mathbf{2}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{\theta }\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{+}\mathbf{2}\mathbf{l}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{l}\mathbf{\text{'}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{\theta }\mathbf{\text{'}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{+}\mathbf{gl}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{sin\theta }\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}\mathbf{0} \end{gathered}$$

Assume g = 1, k = m = 1, and ${\mathbf{l}}_{\mathbf{o}}$= 4. When the system is at rest, $\mathbf{l}\mathbf{=}{\mathbf{l}}_{\mathbf{o}}\mathbf{+}\frac{\mathbf{mg}}{\mathbf{k}}\mathbf{=}\mathbf{5}$.

a. Describe the motion of the pendulum when $\mathbf{l}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{5}\mathbf{.}\mathbf{5}\mathbf{,}\mathbf{l}\mathbf{\text{'}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}\mathbf{,}\mathbf{\theta }\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}\mathbf{,}\mathbf{\theta }\mathbf{\text{'}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}$.

b. When the pendulum is both stretched and given an angular displacement, the motion of the pendulum is more complicated. Using the Runge–Kutta algorithm for systems with h = 0.1 to approximate the solution, sketch the graphs of the length l and the angular displacement u on the interval [0,10] if $\mathbf{l}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{5}\mathbf{.}\mathbf{5}\mathbf{,}\mathbf{l}\mathbf{\text{'}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}\mathbf{,}\mathbf{\theta }\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{5}\mathbf{,}\mathbf{\theta }\mathbf{\text{'}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}$.

In problems 1-6, identify the independent variable, dependent variable, and determine whether the equation is linear or nonlinear.

$\mathbf{6}\mathbf{-}{\mathit{t}}^{\mathbf{3}}\left(\frac{d^{2}v}{d{t}^2}\right)\mathbf{+}\mathbf{3}\mathit{v}\mathbf{-}\mathit{l}\mathit{n}\mathit{t}\mathbf{=}\frac{\mathbf{d}\mathbf{v}}{\mathbf{d}\mathbf{t}}$

Consider the differential equation  $\frac{\mathbf{d}\mathbf{y}}{\mathbf{d}\mathbf{x}}\mathbf{=}\mathit{x}\mathbf{+}\mathit{s}\mathit{i}\mathit{n}\mathbf{ }\mathit{y}$

⦁    A solution curve passes through the point $\left(1,\frac{\pi }{2}\right)$ . What is its slope at this point?

⦁    Argue that every solution curve is increasing for $\mathit{x}\mathbf{>}\mathbf{1}$ .

⦁    Show that the second derivative of every solution satisfies $\frac{{\mathbf{d}}^{\mathbf{2}}\mathbf{y}}{{\mathbf{dx}}^{\mathbf{2}}}\mathbf{=}\mathbf{1}\mathbf{+}\mathbf{x} \mathbf{cos} \mathbf{y}\mathbf{+}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{sin} \mathbf{2}\mathbf{y}\mathbf{.}$ 

⦁    A solution curve passes through (0,0). Prove that this curve has a relative minimum at (0,0).

In problems 1-6, identify the independent variable, dependent variable, and determine whether the equation is linear or nonlinear.

$\mathit{x}\frac{{\mathbf{d}}^{\mathbf{2}}\mathbf{y}}{\mathbf{d}{\mathbf{x}}^{\mathbf{2}}}\mathbf{+}\mathbf{3}\mathit{x}\mathbf{-}\frac{\mathbf{d}\mathbf{y}}{\mathbf{d}\mathbf{x}}\mathbf{=}{\mathit{e}}^{\mathbf{3}\mathbf{x}}$

Consider the differential equation  $\frac{\mathbf{dp}}{\mathbf{dt}}\mathbf{=}\mathbf{p}\mathbf{(}\mathbf{p}\mathbf{-}\mathbf{1}\mathbf{)}\mathbf{(}\mathbf{2}\mathbf{-}\mathbf{p}\mathbf{)}$  for the population p (in thousands) of a certain species at time t.

⦁    Sketch the direction field by using either a computer software package or the method of isoclines.

⦁    If the initial population is 4000 [that is, $p\left(\mathbf{0}\right)=\mathbf{4}$], what can you say about the limiting population  ${\mathbf{lim}}_{\mathbf{t}\to \infty }\mathbf{P}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{?}$

⦁    If $\mathit{p}\left(0\right)\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{7}$ , what is  ${\mathbf{lim}}_{\mathbf{t}\to \infty }\mathbf{P}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{?}$

⦁    If $\mathit{p}\left(0\right)\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{8}$ , what is  ${\mathbf{lim}}_{\mathbf{t}\to \infty }\mathbf{P}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{?}$

⦁    Can a population of 900 ever increase to 1100?

Decide whether the statement made is True or False. The function  $\mathbf{x}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}{\mathbf{t}}^{\mathbf{-}\mathbf{3}}\mathbf{sin} \mathbf{t}\mathbf{+}\mathbf{4}{\mathbf{t}}^{\mathbf{-}\mathbf{3}}$ is a solution to ${\mathbf{t}}^{\mathbf{3}}\frac{\mathbf{dx}}{\mathbf{dt}}\mathbf{=}\mathbf{cos} \mathbf{t}\mathbf{-}\mathbf{3}{\mathbf{t}}^{\mathbf{2}}\mathbf{x}$ .

The motion of a set of particles moving along the x‑axis is governed by the differential equation $\frac{\mathbf{dx}}{\mathbf{dt}}\mathbf{=}{\mathbf{t}}^{\mathbf{3}}\mathbf{-}{\mathbf{x}}^{\mathbf{3}}\mathbf{,}$  where  $\mathbf{x}\left(\mathbf{t}\right)$ denotes the position at time t of the particle.

⦁    If a particle is located at  $\mathit{x}\mathbf{=}\mathbf{1}$ when  $\mathit{t}\mathbf{=}\mathbf{1}$ , what is its velocity at this time?

⦁    Show that the acceleration of a particle is given by  $\frac{{\mathbf{d}}^{\mathbf{2}}\mathbf{x}}{{\mathbf{dt}}^{\mathbf{2}}}\mathbf{=}\mathbf{3}{\mathbf{t}}^{\mathbf{2}}\mathbf{-}\mathbf{3}{\mathbf{t}}^{\mathbf{3}}{\mathbf{x}}^{\mathbf{2}}\mathbf{+}\mathbf{3}{\mathbf{x}}^{\mathbf{5}}\mathbf{.}$

⦁    If a particle is located at  $\mathit{x}\mathbf{=}\mathbf{2}$ when $\mathit{t}\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{5}$, can it reach the location  $\mathit{x}\mathbf{=}\mathbf{1}$ at any later time?

[Hint: ${\mathbf{t}}^{\mathbf{3}}\mathbf{-}{\mathbf{x}}^{\mathbf{3}}\mathbf{=}\mathbf{(}\mathbf{t}\mathbf{-}\mathbf{x}\mathbf{)}\mathbf{(}{\mathbf{t}}^{\mathbf{2}}\mathbf{+}\mathbf{xt}\mathbf{+}{\mathbf{x}}^{\mathbf{2}}\mathbf{)}\mathbf{.}$ ]

Decide whether the statement made is True or False. The function  $\mathbf{y}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}\mathbf{-}\frac{\mathbf{1}}{\mathbf{3}}\mathbf{(}\mathbf{x}\mathbf{+}\mathbf{1}\mathbf{)}$ is a solution to $\frac{\mathbf{dy}}{\mathbf{dx}}\mathbf{=}\frac{\mathbf{y}\mathbf{-}\mathbf{1}}{\mathbf{x}\mathbf{+}\mathbf{3}}$.

Question:8. Determine the convergence set of the given power series.

Let $\varphi \mathbf{\left(}\mathbf{x}\mathbf{\right)}$ denote the solution to the initial value problem  

$\frac{\mathbf{dy}}{\mathbf{dx}}\mathbf{=}\mathbf{x}\mathbf{-}\mathbf{y}\mathbf{,}\mathbf{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{1}$

⦁    Show that  $\varphi \mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}\mathbf{1}\mathbf{-}\varphi \mathbf{\text{'}}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}\mathbf{1}\mathbf{-}\mathbf{x}\mathbf{+}\varphi \mathbf{\left(}\mathbf{x}\mathbf{\right)}$

⦁    Argue that the graph of $\varphi$  is decreasing for x near zero and that as x increases from zero, $\varphi \mathbf{\left(}\mathbf{x}\mathbf{\right)}$ decreases until it crosses the line y = x, where its derivative is zero.

⦁    Let x* be the abscissa of the point where the solution curve  $\mathbf{y}\mathbf{=}\varphi \mathbf{\left(}\mathbf{x}\mathbf{\right)}$ crosses the line $y=x$ .Consider the sign of  $\varphi \mathbf{(}\mathbf{x}\mathbf{*}\mathbf{)}$ and argue that $\varphi$  has a relative minimum at x*.

⦁    What can you say about the graph of  $\mathbf{y}\mathbf{=}\varphi \mathbf{\left(}\mathbf{x}\mathbf{\right)}$ for x > x*?

⦁    Verify that y = x – 1 is a solution to  $\frac{\mathbf{dy}}{\mathbf{dx}}\mathbf{=}\mathbf{x}\mathbf{-}\mathbf{y}$ and explain why the graph of  $\mathbf{y}\mathbf{=}\varphi \mathbf{\left(}\mathbf{x}\mathbf{\right)}$ always stays above the line  $\mathbf{y}\mathbf{=}\mathbf{x}\mathbf{-}\mathbf{1}$.

⦁    Sketch the direction field for  $\frac{\mathbf{dy}}{\mathbf{dx}}\mathbf{=}\mathbf{x}\mathbf{-}\mathbf{y}$ by using the method of isoclines or a computer software package.

⦁    Sketch the solution  $\mathbf{y}\mathbf{=}\varphi \mathbf{\left(}\mathbf{x}\mathbf{\right)}$ using the direction field in part (f).

In Problems 9–20, determine whether the equation is exact.

If it is, then solve it.

$\mathbf{(}\mathbf{2}\mathbf{xy}\mathbf{+}\mathbf{3}\mathbf{)}\mathbf{dx}\mathbf{+}\mathbf{(}{\mathbf{x}}^{\mathbf{2}}\mathbf{-}\mathbf{1}\mathbf{)}\mathbf{dy}\mathbf{=}\mathbf{0}$

Decide whether the statement made is True or False. The relation ${\mathbf{x}}^{\mathbf{2}}\mathbf{+}{\mathbf{y}}^{\mathbf{3}}\mathbf{-}{\mathbf{e}}^{\mathbf{y}}\mathbf{=}\mathbf{1}$  is an implicit solution to $\frac{\mathbf{dy}}{\mathbf{dx}}\mathbf{=}\frac{{\mathbf{e}}^{\mathbf{y}}\mathbf{-}\mathbf{2}\mathbf{x}}{\mathbf{3}{\mathbf{y}}^{\mathbf{2}}}$ .

Find a general solution for the differential equation with x as the independent variable:

$y\text{'}\text{'}\text{'}+3y\text{'}\text{'}-4y\text{'}-6y=0$

Decide whether the statement made is True or False. The relation $\mathbf{sin} \mathbf{y}\mathbf{+}{\mathbf{e}}^{\mathbf{y}}\mathbf{=}{\mathbf{x}}^{\mathbf{6}}\mathbf{-}{\mathbf{x}}^{\mathbf{2}}\mathbf{+}\mathbf{x}\mathbf{+}\mathbf{1}$  is an implicit solution to $\frac{\mathbf{dy}}{\mathbf{dx}}\mathbf{=}\frac{\mathbf{6}{\mathbf{x}}^{\mathbf{5}}\mathbf{-}\mathbf{2}\mathbf{x}\mathbf{+}\mathbf{1}}{\mathbf{cos} \mathbf{y}\mathbf{+}{\mathbf{e}}^{\mathbf{y}}}$.

Question 10: In Problems, find the power series expansion  for f(x)+g(x), given the expansions for f(x) and g(x).

10.

Question 11: In Problem, find the first three nonzero terms in the power series expansion for the product f(x) g(x).